Chapter 9.5, Problem 33PPE

a)

To determine

To write: an expression in terms of w for the length of the playground.

Answer to Problem 33PPE

The length of the playground is $2w+l=\text{Length}$ .

Explanation of Solution

Given:

A school is fencing in a rectangular area for a playground.

It plans to enclose the playground using fencing on three sides, as shown.

The school has budgeted enough money for 75ft of fencing material and would like to make a playground with an area of 600 ft².

  

From the diagram, the expression for the ‘ w’ ,

  $\begin{array}{c}w\times l=600\\ w=\frac{600}{l}\end{array}$

So, the width of the playground is $w=\frac{600}{l}$ .

b)

To determine

To evaluate: an equation to find the width ‘ w’ .

Answer to Problem 33PPE

The width of the playground is $26ft\text{ or }12ft$ .

Explanation of Solution

Given:

A school is fencing in a rectangular area for a playground.

It plans to enclose the playground using fencing on three sides, as shown.

The school has budgeted enough money for 75ft of fencing material and would like to make a playground with an area of 600 ft².

  

From the diagram, the expression for the ‘ w’ ,

  $\begin{array}{l}2w+l=\text{Length}\\ 2w+l=75\\ 2w+\frac{600}{w}=75\end{array}$

Now,

Solving the above equation,

  $\begin{array}{l}2w^2+600=75w\\ 2w^2-75w+600=0\end{array}$

Using the quadratic formula,

  $\begin{array}{c}w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{75\pm \sqrt{{75}^2-4\left(2\right)\left(600\right)}}{2\left(2\right)}\\ =\frac{75\pm \sqrt{5625-4800}}{4}\\ =\frac{75\pm \sqrt{825}}{4}\\ =\frac{75\pm 28.72}{4}\\ =25.93,11.57\\ =26,12\end{array}$

So, the width of the playground is $26ft\text{ or }12ft$ .

c)

To determine

To evaluate: the length of the playground.

Answer to Problem 33PPE

The length of the playground is 23. 07ft or 50 ft .

Explanation of Solution

Given:

A school is fencing in a rectangular area for a playground.

It plans to enclose the playground using fencing on three sides, as shown.

The school has budgeted enough money for 75ft of fencing material and would like to make a playground with an area of 600 ft².

  

From the diagram, the expression for the ‘ w’ ,

  $\begin{array}{l}2w+l=\text{Length}\\ 2w+l=75\\ 2w+\frac{600}{w}=75\end{array}$

Now,

Solving the above equation,

  $\begin{array}{l}2w^2+600=75w\\ 2w^2-75w+600=0\end{array}$

Using the quadratic formula,

  $\begin{array}{c}w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{75\pm \sqrt{{75}^2-4\left(2\right)\left(600\right)}}{2\left(2\right)}\\ =\frac{75\pm \sqrt{5625-4800}}{4}\\ =\frac{75\pm \sqrt{825}}{4}\\ =\frac{75\pm 28.72}{4}\\ =25.93,11.57\\ =26,12\end{array}$

So, the width of the playground is $26ft\text{ or }12ft$ .

If width is 26 ft ,

  $\begin{array}{c}\text{length}=\frac{600}{26}\\ =23.07ft\end{array}$

If width is 12 ft ,

  $\begin{array}{c}\text{length}=\frac{600}{12}\\ =50ft\end{array}$

The length of the playground is 23. 07ft or 50 ft .